Re: Olcott KEEPS TRYING to get away with this falsehood

On 5/13/24 6:51 PM, olcott wrote:

On 5/13/2024 4:49 PM, immibis wrote:

On 13/05/24 15:19, olcott wrote:

On 5/12/2024 11:53 PM, immibis wrote:

On 10/05/24 19:49, olcott wrote:

On 5/10/2024 11:12 AM, Richard Damon wrote:

On 5/10/24 11:50 AM, olcott wrote:

On 5/10/2024 9:18 AM, Richard Damon wrote:

On 5/9/24 11:10 PM, olcott wrote:

On 5/9/2024 9:31 PM, Richard Damon wrote:

On 5/9/24 11:38 AM, olcott wrote:

On 5/8/2024 8:38 PM, immibis wrote:

On 8/05/24 21:05, olcott wrote:

On 5/8/2024 10:13 AM, Mike Terry wrote:

On 08/05/2024 14:01, olcott wrote:

On 5/8/2024 3:59 AM, Mikko wrote:

On 2024-05-07 19:05:54 +0000, olcott said:
>

On 5/7/2024 1:54 PM, Fred. Zwarts wrote:

Op 07.mei.2024 om 17:40 schreef olcott:

On 5/7/2024 6:18 AM, Richard Damon wrote:

On 5/7/24 3:30 AM, Mikko wrote:

On 2024-05-06 18:28:37 +0000, olcott said:
>

On 5/6/2024 11:19 AM, Mikko wrote:

On 2024-05-05 17:02:25 +0000, olcott said:
>

The x86utm operating system: https://github.com/plolcott/x86utm enables
one C function to execute another C function in debug step mode.
Simulating Termination analyzer H simulates the x86 machine code of its
input (using libx86emu) in debug step mode until it correctly matches a
correct non-halting behavior pattern proving that its input will never
stop running unless aborted.
>
Can D correctly simulated by H terminate normally?
00 int H(ptr x, ptr x)  // ptr is pointer to int function
01 int D(ptr x)
02 {
03   int Halt_Status = H(x, x);
04   if (Halt_Status)
05     HERE: goto HERE;
06   return Halt_Status;
07 }
08
09 int main()
10 {
11   H(D,D);
12 }
>
*Execution Trace*
Line 11: main() invokes H(D,D);
>
*keeps repeating* (unless aborted)
Line 03: simulated D(D) invokes simulated H(D,D) that simulates D(D)
>
*Simulation invariant*
D correctly simulated by H cannot possibly reach past its own line 03.
>
The above execution trace proves that (for every H/D pair of the
infinite set of H/D pairs) each D(D) simulated by the H that this D(D)
calls cannot possibly reach past its own line 03.

>
When you say "every H/D pair" you should specify which set of pairs
you are talking about. As you don't, your words don't mean anything.
>

>
Every H/D pair in the universe where D(D) is simulated by the
same H(D,D) that D(D) calls. This involves 1 to ∞ steps of D
and also includes zero to ∞ recursive simulations where H
H simulates itself simulating D(D).

>
"In the universe" is not a set. In typical set theories like ZFC there
is no universal set.

>

>
This template defines an infinite set of finite string H/D pairs where each D(D) that is simulated by H(D,D) also calls this same H(D,D).
>
These H/D pairs can be enumerated by the one to ∞ simulated steps of D and involve zero to ∞ recursive simulations of H simulating itself simulating D(D). Every time Lines 1,2,3 are simulated again defines
one more level of recursive simulation.
>
1st element of H/D pairs 1 step  of D  is simulated by H
2nd element of H/D pairs 2 steps of D are simulated by H
3rd element of H/D pairs 3 steps of D are simulated by H
>
4th element of H/D pairs 4 steps of D are simulated by H
this begins the first recursive simulation at line 01
>
5th element of H/D pairs 5 steps of D are simulated by
next step of the first recursive simulation at line 02
>
6th element of H/D pairs 6 steps of D are simulated by
last step of the first recursive simulation at line 03
>
7th element of H/D pairs 7 steps of D are simulated by H
this begins the second recursive simulation at line 01

>
Is this the definition of the infinite set of H? We can think of many more simulations that only these.

>
This template defines an infinite set of finite string H/D pairs where
each D(D) that is simulated by H(D,D) also calls this same H(D,D).
>
No-one can possibly show one element of this set where D(D) reaches
past its own line 03.

>
If H is a decider of any kind then the D build from it reaches its line
4 as numberd above. Whether the simulation of D by H reaches that line
is another question.
>

>
*My fully operational code proves otherwise*
>
I seems like you guys don't have a clue about how infinite
recursion works. You can run the code and see that I am correct.
>
I have one concrete instance as fully operational code.
https://github.com/plolcott/x86utm/blob/master/Halt7.c
line 555 u32 HH(ptr P, ptr I) its input in on
line 932 int DD(int (*x)())

>
HH is completely broken - it uses a global variable which is allows HH to detect whether it is the outer HH or a nested (simulated) HH. As a result, the nested HH behaves completely differently to the outer HH - I mean /completely/ differently: it goes through a totally separate "I am called in nested mode" code path!
>

>
The encoding of HH is not the pure function that it needs to be to
be a computable function.
>
*Maybe you can settle this*
>
The disagreement is entirely over an enormously much simpler thing.
The disagreement is that Richard says that a D simulated by H could
reach past its own line 03 and halt.

>
Here's the proof:
>
1. A simulation always produces an identical execution trace to the direct execution.

>
*When pathological self-reference is involved this is counter-factual*
That no one can possibly show the steps of how D simulated by H possibly
reach line 06 of H proves this.

>
>
>

>
Richard tried to get away with D never simulated by H as an example
of D simulated by H:

>
Nope, you are looking at the WRONG message, and I have told you this multiple times.

>
Message-ID: <v0ummt$2qov3$2@i2pn2.org>
*When you interpret*
On 5/1/2024 7:28 PM, Richard Damon wrote:
 > On 5/1/24 11:51 AM, olcott wrote:
*Every D simulated by H that cannot possibly*
*stop running unless aborted by H*
>
as *D NEVER simulated by H*
>
you have shown a reckless disregard for the truth
that would win a defamation case.
>

>
My H simulated 0 steps of D, of which was ALL of the steps it simulated correctly.

>
*THAT DOES NOT MEET THE SPEC*

>
You haven't GIVEN a defined SPEC.
>
The only definition within Computation Theory, which is the space you started in, and claim to get to, doesn't have "aborted" simulations, so you don't have a defintion of what simulatioin actually means, other than doing something that tells you something about the behavior of what is simulated.
>
My H does that, by aborting its "simulation" in shows that THIS H did not simulate its input to a final state.
>
Just the same result that you partial set of H's showed.
>

*THAT DOES NOT MEET THE SPEC*
*THAT DOES NOT MEET THE SPEC*
*THAT DOES NOT MEET THE SPEC*
>
*Every D simulated by H that cannot possibly*
*stop running unless aborted by H*
>
*Every D simulated by H that cannot possibly*
*stop running unless aborted by H*
>
*Every D simulated by H that cannot possibly*
*stop running unless aborted by H*
>
*Every D simulated by H that cannot possibly*
*stop running unless aborted by H*
>
*Every D simulated by H that cannot possibly*
*stop running unless aborted by H*
>
>

>
Right, and simulating zero steps correctly and them aborting means H

>
*cannot possibly stop running unless aborted is not met*
*cannot possibly stop running unless aborted is not met*
*cannot possibly stop running unless aborted is not met*
*cannot possibly stop running unless aborted is not met*
*cannot possibly stop running unless aborted is not met*

>
If H never simulates any steps and always aborts then its input cannot possibly stop running unless aborted because its input never has the chance to execute a return instruction before the abort.

>
The instant that the executed H(D,D) stops simulating its input all simulations immediately stop running.
>

because they were aborted.

 If my memory is correct Richard said that some simulated H return a
value to their simulated D even after the outermost D is no longer
simulated.

No, the simulated H's will return a value to the D that called them.

 The key falsehood is that he claims D simulated by H can reach past its
own line 03. I am sure that he means a D that is NOT simulated by H.
 

I proved it in a post that you seem to be ignoring. Since you have been told about it, and do not believe your version enough to take the put up or shut up challenge, you are just demonstrating that you are nothing but a pathological liar.
Try to refute my example, If you need to admit that people have said things that you have ignored, and thus it is just a LIE for you to claim that no one has refuted you, the do so and I will point it out.
Remember, if you do this, if you EVER try to claim that no one has refuted your statement, you are admitting that you are just being a liar again.